Electromagnetic induction > Magnetic flux conservation. Superconductors in magnetic field > Problem 11.5.21

Statement

$11.5.21.$Solve the problem $11.5.20$ if the mass of the first projectile is $m_1$ and the second is $m_2$, and the velocity of the projectiles is $v_1$ and $v_2$, respectively ($v_1 > v_2$).

Solution

The boundary condition will be the same as in the previous problem, but the left side of the inequality will change.

The velocity of the center of mass of the system:

$$\upsilon_C = \frac{m_1\upsilon_1 + m_2\upsilon_2}{m_1 + m_2}$$

Hence the velocity of our projectiles:

$$\upsilon_{1_C} = \frac{m_2(\upsilon_1 - \upsilon_2)}{m_1 + m_2}$$
$$\upsilon_{2_C} = -\frac{m_1(\upsilon_1 - \upsilon_2)}{m_1 + m_2}$$

Now let's calculate the total kinetic energy in the SCM, it will go to the left side of the inequality:

$$E_K = \frac{(\upsilon_1 - \upsilon_2)^2}{2(\frac{1}{m_1}+\frac{1}{m_2})}$$

Finite velocities in the case of $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} < \frac{2B^2 l S^2(3s - 2S)}{\mu_0 (S - s)(2S - s)}$：

They are obtained by changing the sign of the speeds in the CM system and adding its speed:

$${\upsilon_1}' = \frac{2m_2\upsilon_2 + (m_1 - m_2)\upsilon_1}{m_1 + m_2}$$
$${\upsilon_2}' = \frac{2m_1\upsilon_1 + (m_2 - m_1)\upsilon_2}{m_1 + m_2}$$

For the case of $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} > \frac{2B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$ speeds will remain the same:

$${\upsilon_1}' = \upsilon_1$$

$${\upsilon_2}' = \upsilon_2$$

Answer

For:

$$\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} < \frac{2B^2 l S^2(3s - 2S)}{\mu_0 (S - s)(2S - s)}$$

$${\upsilon_1}' = \frac{2m_2\upsilon_2 + (m_1 - m_2)\upsilon_1}{m_1 + m_2}$$
$${\upsilon_2}' = \frac{2m_1\upsilon_1 + (m_2 - m_1)\upsilon_2}{m_1 + m_2}$$

For the case of $\frac{(\upsilon_1 - \upsilon_2)^2}{(\frac{1}{m_1}+\frac{1}{m_2})} > \frac{2B^2 l S^2(3s - 2S)}{\mu_0(S - s)(2S - s)}$ speeds will remain the same:

$${\upsilon_1}' = \upsilon_1$$

$${\upsilon_2}' = \upsilon_2$$